Download Fundamental Law of Electrostatics

Magnetism
The Magnetic Force


 
F  qE  qv  B
B
x x x x x x
x x x x x x
v
x x x x x x
F q
B
B

v

 q
F
v
q
F=0
Today...
• Magnetic forces: The Lorentz Force equation
• Motion of charged particle in a Constant Magnetic
Field.
Magnetic Fields
• We saw last lecture that some substances,
particularly iron, possess a property we call
magnetism that exerts forces on other magnetic
materials
• We also saw that single magnetic charges
(magnetic monopoles) did not exist
• We saw that magnetic fields, shown up by iron
filings look similar to electric dipole fields
• Also that magnetic fields seem to be associated
with moving charges
• What is this "magnetic force"? How is it related to
and distinguished from the "electric" force?
Magnetic Forces
•
Consider a positive charge q
moving in the field Bof a
magnet with velocity v,
experimentally we find:
1. If q moves in the +z direction
and the field points in the +y
direction then the force Fis
in the –x direction. The force
is proportional to the velocity
and the field
2. If q moves in the +x direction
the force is in the +z direction,
again proportional to B and v
Magnetic Forces
3. If q moves in the +y direction
there is no force
4. If q is at rest there is no force
5. The force is proportional to B
6. The force is proportional to
the sign and magnitude of q
The magnetic force Fon a moving charge is proportional
to q, vp and B, where vp is the velocity component
perpendicular to the field, while the direction of Fis
perpendicular to both Band v and depends on the sign
of q
F  qv  B
Lorentz Force
• We can add the effect of an Electric Field and get
the “Lorentz Force”
• The force F on a charge q moving with velocity v
through a region of space with electric field E and
magnetic field B is given by:


 
F  qE  qv  B
B
x x x x x x
B

x x x x x x
v
x x x x x x
q
F
v

 q
F
B
v
q
F=0
Reminder: The Cross Product
• The cross (vector) product of two vectors is a
third vector
– Remember the dot (scalar) product multiplied two vectors
to produce a scalar
B
AXB=C

• The magnitude of C is given by:
A
C = AB sin 
C
• The direction of C is perpendicular to the
plane defined by A and B, and in the
direction defined by the right hand rule,
rotating from A to B.
UIUC
Reminder: The Cross Product
• Cartesian components of the cross product:
B
C=AXB
CX = AY BZ - BY AZ
CY = AZ BX - BZ AX
CZ = AX BY - BX AY
A
Note: B X A = - A X B
C
•Drawing 3-dimensional vectors, conventionally
– a vector going into the slide
– a vector coming out of the slide
Motion in a magnetic field
F  qv  B
Three points are arranged in a uniform
magnetic field. The B field points into the
screen. Consider the force on a positively
charged particle in the following conditions
1) It is located at point A and is stationary.
x
z
•v=0 The magnetic force is zero
2) The positive charge moves from point A toward B.
•v in x direction, B in z, RH rule says F in –y
•The direction of the magnetic force on the particle is to the left
3) The positive charge moves from point A toward C.
•Rotate our x axis to be along the direction A-C
• F will be perpendicular to that line and upwards
y
Question 1
• Two protons each move at speed v in the x-y
plane (as shown in the diagram) in a region
of space which contains a constant B field in
the -z-direction. Ignore the interaction
between the two protons.
– What is the relation between the
magnitudes of the forces on the two
protons?
(a) F1 < F2
(b) F1 = F2
y
1
v
B
2
z
(c) F1 > F2
v
x
Question 1
1. a
2. b
3. c
33%
33%
33%
c
b
a
0
of
300
15
Question 1
• Two protons each move at speed v in the x-y
plane (as shown in the diagram) in a region
of space which contains a constant B field in
the -z-direction. Ignore the interaction
between the two protons.
– What is the relation between the
magnitudes of the forces on the two
protons?
(a) F1 < F2
(b) F1 = F2
y
v
1
B
v
2
x
z
(c) F1 > F2
•The magnetic force is given by:
F  qv  B
F  qvB sin 
• In both cases the angle between v and B is 90°
• Therefore F1 = F2.
Question 2
• Two protons each move at speed v in the x-y
plane (as shown in the diagram) in a region
of space which contains a constant B field in
the -z-direction. Ignore the interaction
between the two protons.
– What is F2x ,the x-component of the force
on the second proton?
(a) F2x < 0
(b) F2x = 0
y
1
v
B
2
v
z
(c) F2x > 0
x
Question 2
1. a
2. b
3. c
33%
33%
33%
c
b
a
0
of
300
15
Question 2
• Two protons each move at speed v in the x-y
plane (as shown in the diagram) in a region
of space which contains a constant B field in
the -z-direction. Ignore the interaction
between the two protons.
– What is F2x ,the x-component of the force
on the second proton?
(a) F2x < 0
(b) F2x = 0
y
1
v
B
2
v
z
x
(c) F2x > 0
•To determine the direction of the force, we use the
right-hand rule.

 
F  qv  B
•The directions of the forces are shown in the diagram
F2x < 0
Question 3
• Two protons each move at speed v in the x-y
plane (as shown in the diagram) in a region
of space which contains a constant B field in
the -z-direction. Ignore the interaction
between the two protons.
– Inside the B field, the speed of each
proton:
(a) decreases
(b) increases
y
1
v
B
2
z
v
x
(c) stays the same
Question 3
1. a
2. b
3. c
33%
33%
33%
c
b
a
0
of
300
15
Question 3
• Two protons each move at speed v in the x-y
plane (as shown in the diagram) in a region
of space which contains a constant B field in
the -z-direction. Ignore the interaction
between the two protons.
– Inside the B field, the speed of each
proton:
(a) decreases
(b) increases
y
1
v
B
2
z
v
x
(c) stays the same
•Although the proton does experience a force (which deflects
it), this is always  to v . Therefore, there is no possibility to
do work
W  F l  Fl cos
• So kinetic energy is constant and
v is constant
Trajectory in a Constant B Field
• Suppose charge q enters B-field with velocity v as
shown below. What will be the path q follows?
x x x x x x x x x x x x
x x x x x x x x x x x vx B
x x x x x x x x x x x x
q
v
F
F
R
• Force is always  to velocity and B.
– Path will be circle. F will be the centripetal force needed to
keep the charge in its circular orbit, radius R.
Radius of Circular Orbit
• Lorentz force:
F  qvB
x x x x x x x x x x x x
• centripetal acc:
v2
a 
R
• Newton's 2nd Law:
F  ma 

x x x x x x x x x x x vx B
x x x x x x x x x x x x
v
F
F q
R
v2
qvB  m
R
mv
R
qB
This is an important result,
with useful experimental
consequences !
Ratio of charge to mass for an
electron
• In 1897 J.J.Thomson
used a cathode ray
tube to measure e/m
for an electron
•Used an electric and magnetic field in opposition
to cancel force and thus deflection of the electron
•Electron accelerated through a voltage V by the
“electron gun” giving it kinetic energy 1 mv 2  eV
2
•Velocity when it enters the fields
v
2eV
m
e/m for an electron
•In an electric field alone the
spot is deflected
•Then apply a magnetic field
at right angles (Force in
opposite direction) until the
deflection is reduced to zero
•Force due to electric field FE  eEkˆ
•Force due to magnetic field FB  ev  B  evBkˆ
•When the fields cancel FE   FB
v
2eV
m
e
E2

m 2VB 2
qEkˆ  qvBkˆ
v
E
B
e
 1.758820  1011 C
kg
m
Measurement of particle energies
•Many experiments using
particles measure their
velocity (energy or
momentum) by measuring
their curvature in a magnetic
field
•Cloud chambers
•Bubble chambers
•Magnetic spectrometers
Preflight 12:
The drawing below shows the top
view of two interconnected chambers.
Each chamber has a unique magnetic
field. A positively charged particle is
fired into chamber 1, and observed to
follow the dashed path shown in the
figure.
5) What is the direction of the magnetic field in chamber 1?
a) Up
b) Down
c) Left
d) Right
e) Into page
f) Out of page
Preflight 12:
6) What is the direction of the magnetic field in chamber 2?
a) Up
b) Down
c) Left
d) Right
e) Into page
f) Out of page
In chamber 1, the velocity is initially up. Since the
particle’s path curves to the right, the force is to the
right as the particle enters the chamber.
Three ways to figure out the direction of B from
this:
1) Put your thumb in the direction of the F (right) and your fingers in the
direction of v (up) The way that your fingers curl is the direction of B.
2) Put your palm in the direction of F (right), and your thumb in the
direction of v (up), your fingers (keep them straight) point in the direction
of B.
3) Keep your thumb, index and middle fingers at right angles from each
other. Your thumb points in the direction of v (up), middle finger points
towards F (right), then the index finger gives the the direction of B (out of
page)
Preflight 12:
8) Compare the magnitude of the magnetic field in chamber 1 to the
magnitude of the magnetic field in chamber 2.
a) B1 > B2
b) B1 = B2
c) B1 < B2
The magnetic force is always perpendicular to v.
The force doesn’t change the magnitude of v, it
only changes the particle’s direction of motion.
The force gives rise to a centripetal acceleration. The radius of curvature is given by:
R
mv
qB
L
Lecture 12, Act 3
• A proton, moving at speed v, enters a
region of space which contains a
constant B field in the -z-direction and is
deflected as shown.
B
v
• Another proton, moving at speed v1 = 2v,
enters the same region of space and is
deflected as shown.
B
B
v1
B
– Compare the work done by the magnetic field
(W for v, W1 for v1) to deflect the protons.
(a) W1 < W
(b) W1 = W
v
(c) W1 > W
v1
L
Lecture 12, Act 3
• A proton, moving at speed v, enters a
region of space which contains a
constant B field in the -z-direction and is
deflected as shown.
v
B
v
• Another proton, moving at speed v1 = 2v,
enters the same region of space and is
deflected as shown.
B
B
v1
v1
B
– Compare the work done by the magnetic field
(W for v, W1 for v1) to deflect the protons.
(a) W1 < W
(b) W1 = W
(c) W1 > W


• Remember that the work done W is defined as: W  F  dx

• Also remember that the magnetic force is always perpendicular to the

 
velocity:
F  qv  B
 
 
• Therefore, the work done is ZERO in each case:
 F  dx   F  vdt  0
Summary
• Lorentz force equation:
   
F  qE  qv  B
– Static B-field does no work
– Velocity-dependent force given by right hand rule
formula
• Next time: magnetic forces and dipoles
Reading assignment: Chapter 28.1 and 28.3
Examples: 28.2, 28.3, and 28.6 through 28.10
The Hall Effect
• Which charges carry current?
• Positive charges moving
• Negative charges moving clockwise
counterclockwise experience
experience upward force
upward force
• Upper plate at higher potential
• Upper plate at lower potential
Equilibrium between electrostatic & magnetic forces:
Fup  qvdrift B
Fdown  qEinduced  q
VH
w
VH  vdrift Bw  "Hall Voltage"
• This type of experiment led to the discovery (E. Hall, 1879) that current in
conductors is carried by negative charges (not always so in semiconductors).
• Can be used as a B-sensor; used in some ABS to detect shaft rotation speed –
ferromagnetic rotating blades interupt the magnetic field  oscillating voltage